Question
If $$\alpha ,\beta $$ are roots of the equation $$6{x^2} + 11x + 3 = 0$$ then
A.
both $${\cos ^{ - 1}}\alpha $$ and $${\cos ^{ - 1}}\beta $$ are real
B.
both $${\operatorname{cosec} ^{ - 1}}\alpha $$ and $${\operatorname{cosec} ^{ - 1}}\beta $$ are real
C.
both $${\cot ^{ - 1}}\alpha $$ and $${\cot ^{ - 1}}\beta $$ are real
D.
None of these
Answer :
both $${\cot ^{ - 1}}\alpha $$ and $${\cot ^{ - 1}}\beta $$ are real
Solution :
$$\eqalign{
& 6{x^2} + 11x + 3 = 0 \cr
& \Rightarrow \,\,x = - \frac{1}{3}, - \frac{3}{2}\,{\text{and}} - 1 < - \frac{1}{3} < 1, - \frac{3}{2} < - 1. \cr} $$
$$\therefore \,\,{\cos^{ - 1}}\left( { - \frac{1}{3}} \right)$$ exists but $${\cos^{ - 1}}\left( { - \frac{3}{2}} \right)$$ does not;
$${\text{cose}}{{\text{c}}^{ - 1}}\left( { - \frac{3}{2}} \right)$$ exists but $${\text{cose}}{{\text{c}}^{ - 1}}\left( { - \frac{1}{3}} \right)$$ does not;
$${\cot ^{ - 1}}\left( { - \frac{1}{3}} \right)$$ and $${\cot ^{ - 1}}\left( { - \frac{3}{2}} \right)$$ exist.